根式方程：结构联想巧用三角代换

解方程组：

$$\{\begin{array}{l}\sqrt{x(1-y)}+\sqrt{y(1-x)}=\frac{1}{2},\\ \sqrt{xy}+\sqrt{(1-x)(1-y)}=\frac{\sqrt{3}}{2}.\end{array}$$

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###分析
由题中结构和 $$0⩽x⩽1,0⩽y⩽1$$，联想到用 $$x={\sin }^2\alpha ,y={\sin }^2\beta$$ 代换解决。

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###解
依题意有 $$0⩽x⩽1,0⩽y⩽1$$，可设 $$x={\sin }^2\alpha ,y={\sin }^2\beta$$，且 $$0⩽\alpha ,\beta ⩽\frac{\pi }{2}$$，

则原方程组可化为
$$\{\begin{array}{l}\sin \alpha \cos \beta +\sin \beta \cos \alpha =\frac{1}{2},\\ \sin \alpha \sin \beta +\cos \alpha \cos \beta =\frac{\sqrt{3}}{2},\end{array}$$
即
$$\{\begin{array}{l}\sin (\alpha +\beta )=\frac{1}{2},\\ \cos (\alpha -\beta )=\frac{\sqrt{3}}{2}.\end{array}$$

因为 $$0⩽\alpha +\beta ⩽\pi ,-\frac{\pi }{2}⩽\alpha -\beta ⩽\frac{\pi }{2}$$，所以
$$\{\begin{array}{l}\alpha +\beta =\frac{\pi }{6}\text{ 或 }\frac{5\pi }{6},\\ \alpha -\beta =\frac{\pi }{6}\text{ 或 }-\frac{\pi }{6},\end{array}$$

故

• $$\{\begin{array}{l}\alpha =\frac{\pi }{6},\\ \beta =0\end{array}$$
• $$\{\begin{array}{l}\alpha =\frac{\pi }{2},\\ \beta =\frac{\pi }{3}\end{array}$$
• $$\{\begin{array}{l}\alpha =0,\\ \beta =\frac{\pi }{6}\end{array}$$
• $$\{\begin{array}{l}\alpha =\frac{\pi }{3},\\ \beta =\frac{\pi }{2}.\end{array}$$

所以原方程组的解：

• $$\{\begin{array}{l}x=\frac{1}{4},\\ y=0\end{array}$$
• $$\{\begin{array}{l}x=1,\\ y=\frac{3}{4}\end{array}$$
• $$\{\begin{array}{l}x=0,\\ y=\frac{1}{4}\end{array}$$
• $$\{\begin{array}{l}x=\frac{3}{4},\\ y=1.\end{array}$$